The Cauchy product of $\sum_{k \ge 0} (-1)^{k}/\sqrt{k+1}$ with itself has $|c_n| \ge 1$ for every $n$, so it diverges

Refuted claim: the Cauchy product of two convergent series of reals converges (The Cauchy product of two series: $c_n = \sum_{k=0}^{n} a_k b_{n-k}$, Series, partial sums, convergence and the sum, divergence, and the tail series).

The witness is a single conditionally convergent series multiplied by itself. Let $(\varepsilon_k)$ be the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$, $s_{\sigma(k)} = -s_k$, which satisfies $|s_k| = 1$, $s \circ e \equiv 1$ and $s \circ o \equiv -1$) and

$a_k \;=\; b_k \;:=\; \frac{\varepsilon_k}{\sqrt{\iota(k+1)}} ,$

so that $\sum a_k$ converges by the alternating series test. Then, as FALSE: the Cauchy product of two convergent series converges establishes,

$|c_n| \;=\; \sum_{k=0}^{n}\frac{1}{\sqrt{\iota(k+1)\,\iota(n-k+1)}} \;\ge\; \frac{2\,\iota(n+1)}{\iota(n+2)} \;\ge\; 1 \qquad (n \in \mathbb{N}),$

so $(c_n)$ does not converge to $0$ and $\sum c_n$ diverges (If a series converges then its terms tend to $0$).

What this counterexample adds to the false statement is the sharp form of the bound: the lower bound $2\iota(n+1)/\iota(n+2) = 2 - 2/\iota(n+2)$ increases to $2$, so $|c_n|$ eventually exceeds every real below $2$. The terms of the product series therefore do not merely fail to tend to $0$; they stay bounded away from it by an amount approaching $2$. Nothing here determines the asymptotic size of $|c_n|$ itself, only this lower bound for it; the divergence is as far from marginal as the bound makes it.

• Why the product cannot cancel. By The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$, $s_{\sigma(k)} = -s_k$, which satisfies $|s_k| = 1$, $s \circ e \equiv 1$ and $s \circ o \equiv -1$ the sign of $a_k b_{n-k}$ is $\varepsilon_k\varepsilon_{n-k} = \varepsilon_n$, the same for every $k \le n$: the antidiagonal of the product array is sign-constant. So all $n+1$ terms of $c_n$ add, and the AM-GM bound (The arithmetic mean, geometric mean inequality, with square roots as in Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$; the positives are $\{x^2 : x \neq 0\}$) shows each is at least $2/\iota(n+2)$.

• Compare the geometric case. In For $|r| < 1$ the Cauchy product of $\sum r^{k}$ with itself is $\sum (k+1) r^{k}$, with sum $1/(1-r)^{2}$ the antidiagonal also has $n+1$ equal terms, but they are $r^{n}$ with $|r| < 1$, so the count $n+1$ is beaten by the decay. Here the terms of the antidiagonal are at least $2/\iota(n+2)$ each, and there are $n+1$ of them, so the count wins.

• This is the same series as in With $a_j = (-1)^{j}/\sqrt{j+1}$ convergent and $b_j = (-1)^{j}$ bounded but not monotone, $\sum a_j b_j = \sum 1/\sqrt{j+1}$ diverges. Its terms are exactly small enough to be summable with alternating signs and not otherwise, which is what makes it the standard witness for both failures.